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-BARSIC- [3]
3 years ago
13

A nonzero polynomial with rational coefficients has all of the numbers [1 sqrt{2}, ; 2 sqrt{3}, ;3 sqrt{4},; dots, ;1000 sqrt{10

01}]as roots. What is the smallest possible degree of such a polynomial?
Mathematics
1 answer:
Colt1911 [192]3 years ago
8 0

Answer:

Its degree can be at least 1970

Step-by-step explanation:

for each root of the form √q, where q is not a square, we have a root -√q. Therefore, we need to find, among the numbers below to 1000, how many sqaures there are.

Since √1000 = 31.6, we have a total of 30 squares:

2², 3², 4², ...., 30², 31²

Each square gives one root and the non squares (there are 1000-30 = 970 of them) gives 2 roots (one for them and one for the opposite). Hence the smallest degree a rational polynomial can have is

970*2 + 30 = 1970

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Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving
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Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

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P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

P(x=0) = ^nC_x P^x (1-P)^n^-^x

P(x=0) = ^5C_0  (0.4)^0 (1-0.4)^5^-^0

P(x=0) = ^5C_0 (0.4)^0 (0.60)^5

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b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

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c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2  (0.4)^2  (0.6)^3

= 0.6826

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

= 1 - [^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2 * (0.4)^2  (0.6)^3]

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e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

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Answer:

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